3.476 \(\int \frac {A+B \tan (c+d x)}{(a+b \tan (c+d x))^{2/3}} \, dx\)

Optimal. Leaf size=357 \[ -\frac {\sqrt {3} (B+i A) \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a-i b}}}{\sqrt {3}}\right )}{2 d (a-i b)^{2/3}}+\frac {\sqrt {3} (-B+i A) \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a+i b}}}{\sqrt {3}}\right )}{2 d (a+i b)^{2/3}}+\frac {3 (B+i A) \log \left (-\sqrt [3]{a+b \tan (c+d x)}+\sqrt [3]{a-i b}\right )}{4 d (a-i b)^{2/3}}-\frac {3 (-B+i A) \log \left (-\sqrt [3]{a+b \tan (c+d x)}+\sqrt [3]{a+i b}\right )}{4 d (a+i b)^{2/3}}-\frac {(-B+i A) \log (\cos (c+d x))}{4 d (a+i b)^{2/3}}+\frac {(B+i A) \log (\cos (c+d x))}{4 d (a-i b)^{2/3}}-\frac {x (A-i B)}{4 (a-i b)^{2/3}}-\frac {x (A+i B)}{4 (a+i b)^{2/3}} \]

[Out]

-1/4*(A-I*B)*x/(a-I*b)^(2/3)-1/4*(A+I*B)*x/(a+I*b)^(2/3)-1/4*(I*A-B)*ln(cos(d*x+c))/(a+I*b)^(2/3)/d+1/4*(I*A+B
)*ln(cos(d*x+c))/(a-I*b)^(2/3)/d+3/4*(I*A+B)*ln((a-I*b)^(1/3)-(a+b*tan(d*x+c))^(1/3))/(a-I*b)^(2/3)/d-3/4*(I*A
-B)*ln((a+I*b)^(1/3)-(a+b*tan(d*x+c))^(1/3))/(a+I*b)^(2/3)/d-1/2*(I*A+B)*arctan(1/3*(1+2*(a+b*tan(d*x+c))^(1/3
)/(a-I*b)^(1/3))*3^(1/2))*3^(1/2)/(a-I*b)^(2/3)/d+1/2*(I*A-B)*arctan(1/3*(1+2*(a+b*tan(d*x+c))^(1/3)/(a+I*b)^(
1/3))*3^(1/2))*3^(1/2)/(a+I*b)^(2/3)/d

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Rubi [A]  time = 0.28, antiderivative size = 357, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3539, 3537, 57, 617, 204, 31} \[ -\frac {\sqrt {3} (B+i A) \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a-i b}}}{\sqrt {3}}\right )}{2 d (a-i b)^{2/3}}+\frac {\sqrt {3} (-B+i A) \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a+i b}}}{\sqrt {3}}\right )}{2 d (a+i b)^{2/3}}+\frac {3 (B+i A) \log \left (-\sqrt [3]{a+b \tan (c+d x)}+\sqrt [3]{a-i b}\right )}{4 d (a-i b)^{2/3}}-\frac {3 (-B+i A) \log \left (-\sqrt [3]{a+b \tan (c+d x)}+\sqrt [3]{a+i b}\right )}{4 d (a+i b)^{2/3}}-\frac {(-B+i A) \log (\cos (c+d x))}{4 d (a+i b)^{2/3}}+\frac {(B+i A) \log (\cos (c+d x))}{4 d (a-i b)^{2/3}}-\frac {x (A-i B)}{4 (a-i b)^{2/3}}-\frac {x (A+i B)}{4 (a+i b)^{2/3}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Tan[c + d*x])/(a + b*Tan[c + d*x])^(2/3),x]

[Out]

-((A - I*B)*x)/(4*(a - I*b)^(2/3)) - ((A + I*B)*x)/(4*(a + I*b)^(2/3)) - (Sqrt[3]*(I*A + B)*ArcTan[(1 + (2*(a
+ b*Tan[c + d*x])^(1/3))/(a - I*b)^(1/3))/Sqrt[3]])/(2*(a - I*b)^(2/3)*d) + (Sqrt[3]*(I*A - B)*ArcTan[(1 + (2*
(a + b*Tan[c + d*x])^(1/3))/(a + I*b)^(1/3))/Sqrt[3]])/(2*(a + I*b)^(2/3)*d) - ((I*A - B)*Log[Cos[c + d*x]])/(
4*(a + I*b)^(2/3)*d) + ((I*A + B)*Log[Cos[c + d*x]])/(4*(a - I*b)^(2/3)*d) + (3*(I*A + B)*Log[(a - I*b)^(1/3)
- (a + b*Tan[c + d*x])^(1/3)])/(4*(a - I*b)^(2/3)*d) - (3*(I*A - B)*Log[(a + I*b)^(1/3) - (a + b*Tan[c + d*x])
^(1/3)])/(4*(a + I*b)^(2/3)*d)

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 57

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/b, 3]}, -Simp[L
og[RemoveContent[a + b*x, x]]/(2*b*q^2), x] + (-Dist[3/(2*b*q), Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x
)^(1/3)], x] - Dist[3/(2*b*q^2), Subst[Int[1/(q - x), x], x, (c + d*x)^(1/3)], x])] /; FreeQ[{a, b, c, d}, x]
&& PosQ[(b*c - a*d)/b]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 3537

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(c*
d)/f, Subst[Int[(a + (b*x)/d)^m/(d^2 + c*x), x], x, d*Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m}, x] &&
NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[c^2 + d^2, 0]

Rule 3539

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(c
+ I*d)/2, Int[(a + b*Tan[e + f*x])^m*(1 - I*Tan[e + f*x]), x], x] + Dist[(c - I*d)/2, Int[(a + b*Tan[e + f*x]
)^m*(1 + I*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0]
&& NeQ[c^2 + d^2, 0] &&  !IntegerQ[m]

Rubi steps

\begin {align*} \int \frac {A+B \tan (c+d x)}{(a+b \tan (c+d x))^{2/3}} \, dx &=\frac {1}{2} (A-i B) \int \frac {1+i \tan (c+d x)}{(a+b \tan (c+d x))^{2/3}} \, dx+\frac {1}{2} (A+i B) \int \frac {1-i \tan (c+d x)}{(a+b \tan (c+d x))^{2/3}} \, dx\\ &=-\frac {(i A-B) \operatorname {Subst}\left (\int \frac {1}{(-1+x) (a+i b x)^{2/3}} \, dx,x,-i \tan (c+d x)\right )}{2 d}+\frac {(i A+B) \operatorname {Subst}\left (\int \frac {1}{(-1+x) (a-i b x)^{2/3}} \, dx,x,i \tan (c+d x)\right )}{2 d}\\ &=-\frac {(A-i B) x}{4 (a-i b)^{2/3}}-\frac {(A+i B) x}{4 (a+i b)^{2/3}}-\frac {(i A-B) \log (\cos (c+d x))}{4 (a+i b)^{2/3} d}+\frac {(i A+B) \log (\cos (c+d x))}{4 (a-i b)^{2/3} d}+\frac {(3 (i A-B)) \operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{a+i b}-x} \, dx,x,\sqrt [3]{a+b \tan (c+d x)}\right )}{4 (a+i b)^{2/3} d}+\frac {(3 (i A-B)) \operatorname {Subst}\left (\int \frac {1}{(a+i b)^{2/3}+\sqrt [3]{a+i b} x+x^2} \, dx,x,\sqrt [3]{a+b \tan (c+d x)}\right )}{4 \sqrt [3]{a+i b} d}-\frac {(3 (i A+B)) \operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{a-i b}-x} \, dx,x,\sqrt [3]{a+b \tan (c+d x)}\right )}{4 (a-i b)^{2/3} d}-\frac {(3 (i A+B)) \operatorname {Subst}\left (\int \frac {1}{(a-i b)^{2/3}+\sqrt [3]{a-i b} x+x^2} \, dx,x,\sqrt [3]{a+b \tan (c+d x)}\right )}{4 \sqrt [3]{a-i b} d}\\ &=-\frac {(A-i B) x}{4 (a-i b)^{2/3}}-\frac {(A+i B) x}{4 (a+i b)^{2/3}}-\frac {(i A-B) \log (\cos (c+d x))}{4 (a+i b)^{2/3} d}+\frac {(i A+B) \log (\cos (c+d x))}{4 (a-i b)^{2/3} d}+\frac {3 (i A+B) \log \left (\sqrt [3]{a-i b}-\sqrt [3]{a+b \tan (c+d x)}\right )}{4 (a-i b)^{2/3} d}-\frac {3 (i A-B) \log \left (\sqrt [3]{a+i b}-\sqrt [3]{a+b \tan (c+d x)}\right )}{4 (a+i b)^{2/3} d}-\frac {(3 (i A-B)) \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a+i b}}\right )}{2 (a+i b)^{2/3} d}+\frac {(3 (i A+B)) \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a-i b}}\right )}{2 (a-i b)^{2/3} d}\\ &=-\frac {(A-i B) x}{4 (a-i b)^{2/3}}-\frac {(A+i B) x}{4 (a+i b)^{2/3}}-\frac {\sqrt {3} (i A+B) \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a-i b}}}{\sqrt {3}}\right )}{2 (a-i b)^{2/3} d}+\frac {\sqrt {3} (i A-B) \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a+i b}}}{\sqrt {3}}\right )}{2 (a+i b)^{2/3} d}-\frac {(i A-B) \log (\cos (c+d x))}{4 (a+i b)^{2/3} d}+\frac {(i A+B) \log (\cos (c+d x))}{4 (a-i b)^{2/3} d}+\frac {3 (i A+B) \log \left (\sqrt [3]{a-i b}-\sqrt [3]{a+b \tan (c+d x)}\right )}{4 (a-i b)^{2/3} d}-\frac {3 (i A-B) \log \left (\sqrt [3]{a+i b}-\sqrt [3]{a+b \tan (c+d x)}\right )}{4 (a+i b)^{2/3} d}\\ \end {align*}

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Mathematica [A]  time = 0.29, size = 305, normalized size = 0.85 \[ \frac {i \left (\frac {(A+i B) \left (2 \sqrt {3} \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a+i b}}}{\sqrt {3}}\right )-2 \log \left (-\sqrt [3]{a+b \tan (c+d x)}+\sqrt [3]{a+i b}\right )+\log \left (\sqrt [3]{a+i b} \sqrt [3]{a+b \tan (c+d x)}+(a+b \tan (c+d x))^{2/3}+(a+i b)^{2/3}\right )\right )}{(a+i b)^{2/3}}-\frac {(A-i B) \left (2 \sqrt {3} \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{a+b \tan (c+d x)}}{\sqrt [3]{a-i b}}}{\sqrt {3}}\right )-2 \log \left (-\sqrt [3]{a+b \tan (c+d x)}+\sqrt [3]{a-i b}\right )+\log \left (\sqrt [3]{a-i b} \sqrt [3]{a+b \tan (c+d x)}+(a+b \tan (c+d x))^{2/3}+(a-i b)^{2/3}\right )\right )}{(a-i b)^{2/3}}\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Tan[c + d*x])/(a + b*Tan[c + d*x])^(2/3),x]

[Out]

((I/4)*(-(((A - I*B)*(2*Sqrt[3]*ArcTan[(1 + (2*(a + b*Tan[c + d*x])^(1/3))/(a - I*b)^(1/3))/Sqrt[3]] - 2*Log[(
a - I*b)^(1/3) - (a + b*Tan[c + d*x])^(1/3)] + Log[(a - I*b)^(2/3) + (a - I*b)^(1/3)*(a + b*Tan[c + d*x])^(1/3
) + (a + b*Tan[c + d*x])^(2/3)]))/(a - I*b)^(2/3)) + ((A + I*B)*(2*Sqrt[3]*ArcTan[(1 + (2*(a + b*Tan[c + d*x])
^(1/3))/(a + I*b)^(1/3))/Sqrt[3]] - 2*Log[(a + I*b)^(1/3) - (a + b*Tan[c + d*x])^(1/3)] + Log[(a + I*b)^(2/3)
+ (a + I*b)^(1/3)*(a + b*Tan[c + d*x])^(1/3) + (a + b*Tan[c + d*x])^(2/3)]))/(a + I*b)^(2/3)))/d

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(d*x+c))/(a+b*tan(d*x+c))^(2/3),x, algorithm="fricas")

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {B \tan \left (d x + c\right ) + A}{{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(d*x+c))/(a+b*tan(d*x+c))^(2/3),x, algorithm="giac")

[Out]

integrate((B*tan(d*x + c) + A)/(b*tan(d*x + c) + a)^(2/3), x)

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maple [C]  time = 0.31, size = 69, normalized size = 0.19 \[ \frac {\munderset {\textit {\_R} =\RootOf \left (\textit {\_Z}^{6}-2 \textit {\_Z}^{3} a +a^{2}+b^{2}\right )}{\sum }\frac {\left (B \,\textit {\_R}^{3}+A b -a B \right ) \ln \left (\left (a +b \tan \left (d x +c \right )\right )^{\frac {1}{3}}-\textit {\_R} \right )}{\textit {\_R}^{5}-\textit {\_R}^{2} a}}{2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*tan(d*x+c))/(a+b*tan(d*x+c))^(2/3),x)

[Out]

1/2/d*sum((B*_R^3+A*b-B*a)/(_R^5-_R^2*a)*ln((a+b*tan(d*x+c))^(1/3)-_R),_R=RootOf(_Z^6-2*_Z^3*a+a^2+b^2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {B \tan \left (d x + c\right ) + A}{{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(d*x+c))/(a+b*tan(d*x+c))^(2/3),x, algorithm="maxima")

[Out]

integrate((B*tan(d*x + c) + A)/(b*tan(d*x + c) + a)^(2/3), x)

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mupad [B]  time = 21.70, size = 4562, normalized size = 12.78 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*tan(c + d*x))/(a + b*tan(c + d*x))^(2/3),x)

[Out]

log((((16*(-B^6*a^2*b^2*d^6)^(1/2) + 8*B^3*a^2*d^3 - 8*B^3*b^2*d^3)/(d^6*(a^2 + b^2)^2))^(1/3)*(1944*a*b^4*(-B
^6*a^2*b^2*d^6)^(1/2) - 1944*B^3*a*b^6*d^3 + 243*B*b^8*d^5*(a + b*tan(c + d*x))^(1/3)*((16*(-B^6*a^2*b^2*d^6)^
(1/2) + 8*B^3*a^2*d^3 - 8*B^3*b^2*d^3)/(d^6*(a^2 + b^2)^2))^(2/3) - 243*B*a^4*b^4*d^5*(a + b*tan(c + d*x))^(1/
3)*((16*(-B^6*a^2*b^2*d^6)^(1/2) + 8*B^3*a^2*d^3 - 8*B^3*b^2*d^3)/(d^6*(a^2 + b^2)^2))^(2/3)))/(4*d^6*(a^2 + b
^2)) + (486*B^4*b^4*(a + b*tan(c + d*x))^(1/3))/d^4)*((((16*B^3*a^2*d^3 - 16*B^3*b^2*d^3)^2/4 - B^6*(64*a^4*d^
6 + 64*b^4*d^6 + 128*a^2*b^2*d^6))^(1/2) + 8*B^3*a^2*d^3 - 8*B^3*b^2*d^3)/(64*(a^4*d^6 + b^4*d^6 + 2*a^2*b^2*d
^6)))^(1/3) + log((486*B^4*b^4*(a + b*tan(c + d*x))^(1/3))/d^4 - ((-(16*(-B^6*a^2*b^2*d^6)^(1/2) - 8*B^3*a^2*d
^3 + 8*B^3*b^2*d^3)/(d^6*(a^2 + b^2)^2))^(1/3)*(1944*a*b^4*(-B^6*a^2*b^2*d^6)^(1/2) + 1944*B^3*a*b^6*d^3 - 243
*B*b^8*d^5*(a + b*tan(c + d*x))^(1/3)*(-(16*(-B^6*a^2*b^2*d^6)^(1/2) - 8*B^3*a^2*d^3 + 8*B^3*b^2*d^3)/(d^6*(a^
2 + b^2)^2))^(2/3) + 243*B*a^4*b^4*d^5*(a + b*tan(c + d*x))^(1/3)*(-(16*(-B^6*a^2*b^2*d^6)^(1/2) - 8*B^3*a^2*d
^3 + 8*B^3*b^2*d^3)/(d^6*(a^2 + b^2)^2))^(2/3)))/(4*d^6*(a^2 + b^2)))*(-(((16*B^3*a^2*d^3 - 16*B^3*b^2*d^3)^2/
4 - B^6*(64*a^4*d^6 + 64*b^4*d^6 + 128*a^2*b^2*d^6))^(1/2) - 8*B^3*a^2*d^3 + 8*B^3*b^2*d^3)/(64*(a^4*d^6 + b^4
*d^6 + 2*a^2*b^2*d^6)))^(1/3) + log(((((1944*a*b^4*(a^2 + b^2)*((8*(-A^6*d^6*(a^2 - b^2)^2)^(1/2) + 16*A^3*a*b
*d^3)/(d^6*(a^2 + b^2)^2))^(1/3) + (7776*A*a*b^5*(a + b*tan(c + d*x))^(1/3))/d)*((8*(-A^6*d^6*(a^2 - b^2)^2)^(
1/2) + 16*A^3*a*b*d^3)/(d^6*(a^2 + b^2)^2))^(2/3))/16 - (972*A^3*b^5)/d^3)*((8*(-A^6*d^6*(a^2 - b^2)^2)^(1/2)
+ 16*A^3*a*b*d^3)/(d^6*(a^2 + b^2)^2))^(1/3))/4 - (486*A^4*b^4*(a + b*tan(c + d*x))^(1/3))/d^4)*(((256*A^6*a^2
*b^2*d^6 - A^6*(64*a^4*d^6 + 64*b^4*d^6 + 128*a^2*b^2*d^6))^(1/2) + 16*A^3*a*b*d^3)/(64*(a^4*d^6 + b^4*d^6 + 2
*a^2*b^2*d^6)))^(1/3) + log(((((1944*a*b^4*(a^2 + b^2)*(-(8*(-A^6*d^6*(a^2 - b^2)^2)^(1/2) - 16*A^3*a*b*d^3)/(
d^6*(a^2 + b^2)^2))^(1/3) + (7776*A*a*b^5*(a + b*tan(c + d*x))^(1/3))/d)*(-(8*(-A^6*d^6*(a^2 - b^2)^2)^(1/2) -
16*A^3*a*b*d^3)/(d^6*(a^2 + b^2)^2))^(2/3))/16 - (972*A^3*b^5)/d^3)*(-(8*(-A^6*d^6*(a^2 - b^2)^2)^(1/2) - 16*
A^3*a*b*d^3)/(d^6*(a^2 + b^2)^2))^(1/3))/4 - (486*A^4*b^4*(a + b*tan(c + d*x))^(1/3))/d^4)*(-((256*A^6*a^2*b^2
*d^6 - A^6*(64*a^4*d^6 + 64*b^4*d^6 + 128*a^2*b^2*d^6))^(1/2) - 16*A^3*a*b*d^3)/(64*(a^4*d^6 + b^4*d^6 + 2*a^2
*b^2*d^6)))^(1/3) + log((486*B^4*b^4*(a + b*tan(c + d*x))^(1/3))/d^4 - ((3^(1/2)*1i - 1)*(((3^(1/2)*1i + 1)*(9
72*a*b^4*(3^(1/2)*1i - 1)*(a^2 + b^2)*((16*(-B^6*a^2*b^2*d^6)^(1/2) + 8*B^3*a^2*d^3 - 8*B^3*b^2*d^3)/(d^6*(a^2
+ b^2)^2))^(1/3) - (3888*B*b^4*(a^2 - b^2)*(a + b*tan(c + d*x))^(1/3))/d)*((16*(-B^6*a^2*b^2*d^6)^(1/2) + 8*B
^3*a^2*d^3 - 8*B^3*b^2*d^3)/(d^6*(a^2 + b^2)^2))^(2/3))/32 + (972*B^3*a*b^4)/d^3)*((16*(-B^6*a^2*b^2*d^6)^(1/2
) + 8*B^3*a^2*d^3 - 8*B^3*b^2*d^3)/(d^6*(a^2 + b^2)^2))^(1/3))/8)*((3^(1/2)*(((-256*B^6*a^2*b^2*d^6)^(1/2) + 8
*B^3*a^2*d^3 - 8*B^3*b^2*d^3)/(64*a^4*d^6 + 64*b^4*d^6 + 128*a^2*b^2*d^6))^(1/3)*1i)/2 - (((-256*B^6*a^2*b^2*d
^6)^(1/2) + 8*B^3*a^2*d^3 - 8*B^3*b^2*d^3)/(64*a^4*d^6 + 64*b^4*d^6 + 128*a^2*b^2*d^6))^(1/3)/2) + log((486*B^
4*b^4*(a + b*tan(c + d*x))^(1/3))/d^4 - ((3^(1/2)*1i - 1)*(((3^(1/2)*1i + 1)*(972*a*b^4*(3^(1/2)*1i - 1)*(a^2
+ b^2)*(-(16*(-B^6*a^2*b^2*d^6)^(1/2) - 8*B^3*a^2*d^3 + 8*B^3*b^2*d^3)/(d^6*(a^2 + b^2)^2))^(1/3) - (3888*B*b^
4*(a^2 - b^2)*(a + b*tan(c + d*x))^(1/3))/d)*(-(16*(-B^6*a^2*b^2*d^6)^(1/2) - 8*B^3*a^2*d^3 + 8*B^3*b^2*d^3)/(
d^6*(a^2 + b^2)^2))^(2/3))/32 + (972*B^3*a*b^4)/d^3)*(-(16*(-B^6*a^2*b^2*d^6)^(1/2) - 8*B^3*a^2*d^3 + 8*B^3*b^
2*d^3)/(d^6*(a^2 + b^2)^2))^(1/3))/8)*((3^(1/2)*(-((-256*B^6*a^2*b^2*d^6)^(1/2) - 8*B^3*a^2*d^3 + 8*B^3*b^2*d^
3)/(64*a^4*d^6 + 64*b^4*d^6 + 128*a^2*b^2*d^6))^(1/3)*1i)/2 - (-((-256*B^6*a^2*b^2*d^6)^(1/2) - 8*B^3*a^2*d^3
+ 8*B^3*b^2*d^3)/(64*a^4*d^6 + 64*b^4*d^6 + 128*a^2*b^2*d^6))^(1/3)/2) - log(- (((3^(1/2)*1i)/2 + 1/2)*((((3^(
1/2)*1i)/2 - 1/2)*(1944*a*b^4*((3^(1/2)*1i)/2 + 1/2)*(a^2 + b^2)*((16*(-B^6*a^2*b^2*d^6)^(1/2) + 8*B^3*a^2*d^3
- 8*B^3*b^2*d^3)/(d^6*(a^2 + b^2)^2))^(1/3) + (3888*B*b^4*(a^2 - b^2)*(a + b*tan(c + d*x))^(1/3))/d)*((16*(-B
^6*a^2*b^2*d^6)^(1/2) + 8*B^3*a^2*d^3 - 8*B^3*b^2*d^3)/(d^6*(a^2 + b^2)^2))^(2/3))/16 + (972*B^3*a*b^4)/d^3)*(
(16*(-B^6*a^2*b^2*d^6)^(1/2) + 8*B^3*a^2*d^3 - 8*B^3*b^2*d^3)/(d^6*(a^2 + b^2)^2))^(1/3))/4 - (486*B^4*b^4*(a
+ b*tan(c + d*x))^(1/3))/d^4)*((3^(1/2)*1i)/2 + 1/2)*((((16*B^3*a^2*d^3 - 16*B^3*b^2*d^3)^2/4 - B^6*(64*a^4*d^
6 + 64*b^4*d^6 + 128*a^2*b^2*d^6))^(1/2) + 8*B^3*a^2*d^3 - 8*B^3*b^2*d^3)/(64*(a^4*d^6 + b^4*d^6 + 2*a^2*b^2*d
^6)))^(1/3) - log(- (((3^(1/2)*1i)/2 + 1/2)*((((3^(1/2)*1i)/2 - 1/2)*(1944*a*b^4*((3^(1/2)*1i)/2 + 1/2)*(a^2 +
b^2)*(-(16*(-B^6*a^2*b^2*d^6)^(1/2) - 8*B^3*a^2*d^3 + 8*B^3*b^2*d^3)/(d^6*(a^2 + b^2)^2))^(1/3) + (3888*B*b^4
*(a^2 - b^2)*(a + b*tan(c + d*x))^(1/3))/d)*(-(16*(-B^6*a^2*b^2*d^6)^(1/2) - 8*B^3*a^2*d^3 + 8*B^3*b^2*d^3)/(d
^6*(a^2 + b^2)^2))^(2/3))/16 + (972*B^3*a*b^4)/d^3)*(-(16*(-B^6*a^2*b^2*d^6)^(1/2) - 8*B^3*a^2*d^3 + 8*B^3*b^2
*d^3)/(d^6*(a^2 + b^2)^2))^(1/3))/4 - (486*B^4*b^4*(a + b*tan(c + d*x))^(1/3))/d^4)*((3^(1/2)*1i)/2 + 1/2)*(-(
((16*B^3*a^2*d^3 - 16*B^3*b^2*d^3)^2/4 - B^6*(64*a^4*d^6 + 64*b^4*d^6 + 128*a^2*b^2*d^6))^(1/2) - 8*B^3*a^2*d^
3 + 8*B^3*b^2*d^3)/(64*(a^4*d^6 + b^4*d^6 + 2*a^2*b^2*d^6)))^(1/3) + log((((3^(1/2)*1i)/2 - 1/2)*((972*A^3*b^5
)/d^3 + (((3^(1/2)*1i)/2 + 1/2)*(1944*a*b^4*((3^(1/2)*1i)/2 - 1/2)*(a^2 + b^2)*((8*(-A^6*d^6*(a^2 - b^2)^2)^(1
/2) + 16*A^3*a*b*d^3)/(d^6*(a^2 + b^2)^2))^(1/3) + (7776*A*a*b^5*(a + b*tan(c + d*x))^(1/3))/d)*((8*(-A^6*d^6*
(a^2 - b^2)^2)^(1/2) + 16*A^3*a*b*d^3)/(d^6*(a^2 + b^2)^2))^(2/3))/16)*((8*(-A^6*d^6*(a^2 - b^2)^2)^(1/2) + 16
*A^3*a*b*d^3)/(d^6*(a^2 + b^2)^2))^(1/3))/4 + (486*A^4*b^4*(a + b*tan(c + d*x))^(1/3))/d^4)*((3^(1/2)*1i)/2 -
1/2)*(((256*A^6*a^2*b^2*d^6 - A^6*(64*a^4*d^6 + 64*b^4*d^6 + 128*a^2*b^2*d^6))^(1/2) + 16*A^3*a*b*d^3)/(64*(a^
4*d^6 + b^4*d^6 + 2*a^2*b^2*d^6)))^(1/3) - log((((3^(1/2)*1i)/2 + 1/2)*((972*A^3*b^5)/d^3 + (((3^(1/2)*1i)/2 -
1/2)*(1944*a*b^4*((3^(1/2)*1i)/2 + 1/2)*(a^2 + b^2)*((8*(-A^6*d^6*(a^2 - b^2)^2)^(1/2) + 16*A^3*a*b*d^3)/(d^6
*(a^2 + b^2)^2))^(1/3) - (7776*A*a*b^5*(a + b*tan(c + d*x))^(1/3))/d)*((8*(-A^6*d^6*(a^2 - b^2)^2)^(1/2) + 16*
A^3*a*b*d^3)/(d^6*(a^2 + b^2)^2))^(2/3))/16)*((8*(-A^6*d^6*(a^2 - b^2)^2)^(1/2) + 16*A^3*a*b*d^3)/(d^6*(a^2 +
b^2)^2))^(1/3))/4 - (486*A^4*b^4*(a + b*tan(c + d*x))^(1/3))/d^4)*((3^(1/2)*1i)/2 + 1/2)*(((256*A^6*a^2*b^2*d^
6 - A^6*(64*a^4*d^6 + 64*b^4*d^6 + 128*a^2*b^2*d^6))^(1/2) + 16*A^3*a*b*d^3)/(64*(a^4*d^6 + b^4*d^6 + 2*a^2*b^
2*d^6)))^(1/3) + log((((3^(1/2)*1i)/2 - 1/2)*((972*A^3*b^5)/d^3 + (((3^(1/2)*1i)/2 + 1/2)*(1944*a*b^4*((3^(1/2
)*1i)/2 - 1/2)*(a^2 + b^2)*(-(8*(-A^6*d^6*(a^2 - b^2)^2)^(1/2) - 16*A^3*a*b*d^3)/(d^6*(a^2 + b^2)^2))^(1/3) +
(7776*A*a*b^5*(a + b*tan(c + d*x))^(1/3))/d)*(-(8*(-A^6*d^6*(a^2 - b^2)^2)^(1/2) - 16*A^3*a*b*d^3)/(d^6*(a^2 +
b^2)^2))^(2/3))/16)*(-(8*(-A^6*d^6*(a^2 - b^2)^2)^(1/2) - 16*A^3*a*b*d^3)/(d^6*(a^2 + b^2)^2))^(1/3))/4 + (48
6*A^4*b^4*(a + b*tan(c + d*x))^(1/3))/d^4)*((3^(1/2)*1i)/2 - 1/2)*(-((256*A^6*a^2*b^2*d^6 - A^6*(64*a^4*d^6 +
64*b^4*d^6 + 128*a^2*b^2*d^6))^(1/2) - 16*A^3*a*b*d^3)/(64*(a^4*d^6 + b^4*d^6 + 2*a^2*b^2*d^6)))^(1/3) - log((
((3^(1/2)*1i)/2 + 1/2)*((972*A^3*b^5)/d^3 + (((3^(1/2)*1i)/2 - 1/2)*(1944*a*b^4*((3^(1/2)*1i)/2 + 1/2)*(a^2 +
b^2)*(-(8*(-A^6*d^6*(a^2 - b^2)^2)^(1/2) - 16*A^3*a*b*d^3)/(d^6*(a^2 + b^2)^2))^(1/3) - (7776*A*a*b^5*(a + b*t
an(c + d*x))^(1/3))/d)*(-(8*(-A^6*d^6*(a^2 - b^2)^2)^(1/2) - 16*A^3*a*b*d^3)/(d^6*(a^2 + b^2)^2))^(2/3))/16)*(
-(8*(-A^6*d^6*(a^2 - b^2)^2)^(1/2) - 16*A^3*a*b*d^3)/(d^6*(a^2 + b^2)^2))^(1/3))/4 - (486*A^4*b^4*(a + b*tan(c
+ d*x))^(1/3))/d^4)*((3^(1/2)*1i)/2 + 1/2)*(-((256*A^6*a^2*b^2*d^6 - A^6*(64*a^4*d^6 + 64*b^4*d^6 + 128*a^2*b
^2*d^6))^(1/2) - 16*A^3*a*b*d^3)/(64*(a^4*d^6 + b^4*d^6 + 2*a^2*b^2*d^6)))^(1/3)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {A + B \tan {\left (c + d x \right )}}{\left (a + b \tan {\left (c + d x \right )}\right )^{\frac {2}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(d*x+c))/(a+b*tan(d*x+c))**(2/3),x)

[Out]

Integral((A + B*tan(c + d*x))/(a + b*tan(c + d*x))**(2/3), x)

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